如何阻止Scrapy CrawlSpider访问超过要求的URL？

from scrapy.spiders import Rule
from scrapy.linkextractors import LinkExtractor
from scrapy.spiders import CrawlSpider

class MySpider(CrawlSpider):
    name = 'myspider'
    allowed_domains = ['example.com']
    start_urls = ['http://www.example.com']

    rules = (
        Rule(LinkExtractor(allow=r'/page/\d+'), callback='parse_item', follow=True),
    )

    def parse_item(self, response):
        # 处理爬取到的数据
        pass

from scrapy.spiders import CrawlSpider

class MySpider(CrawlSpider):
    name = 'myspider'
    allowed_domains = ['example.com']
    start_urls = ['http://www.example.com']

    def parse(self, response):
        if not response.url.endswith('/page/1'):
            return  # 不符合要求的URL直接返回，不进行爬取

        # 继续处理爬取到的数据
        pass

from scrapy import signals
from scrapy.exceptions import IgnoreRequest
import re

class URLFilterMiddleware:
    def __init__(self, allowed_url_patterns):
        self.allowed_url_patterns = allowed_url_patterns

    @classmethod
    def from_crawler(cls, crawler):
        allowed_url_patterns = getattr(crawler.spider, 'allowed_url_patterns', [])
        return cls(allowed_url_patterns)

    def process_request(self, request, spider):
        url = request.url
        for pattern in self.allowed_url_patterns:
            if re.match(pattern, url):
                return None  # URL匹配，继续请求
        raise IgnoreRequest  # URL不匹配，忽略请求

DOWNLOADER_MIDDLEWARES = {
    'myproject.middlewares.URLFilterMiddleware': 543,
}

from scrapy.spiders import CrawlSpider

class MySpider(CrawlSpider):
    name = 'myspider'
    allowed_domains = ['example.com']
    start_urls = ['http://www.example.com']
    allowed_url_patterns = [
        r'^http://www\.example\.com/page/\d+$',  # 允许匹配的URL规则
    ]

    def parse(self, response):
        # 处理爬取到的数据
        pass